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I’m really wondering how can decreasing the temperature refer to an endothermic reaction.

I faced this when I was calculating the heat of a reaction Q in a constant-pressure calorimeter.

What I believe is that when something becomes cooler, then it has released some amount of heat.

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Let’s say that you have a beaker of water. You measure the water’s temperature using a thermometer, and you determine that it is 25 °C. Now you add two reactants to the beaker, and mix them together well until they are completely reacted. If the reaction between these two reactants is endothermic, then the chemicals will absorb heat from the water. If you were to now remeasure the water’s temperature, you might find that it was something like 23 °C. This is because the heat that was absorbed by the chemicals was lost by the water.

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  • $\begingroup$ What if the reaction is endothermic and exergonic? The opposite reaction would be exothermic. So in the equilibrium products are favor. Therefore shouldn't the temperature increase? $\endgroup$ – ado sar Apr 30 '20 at 16:51
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In case of endothermic reactions, products possess more energy than substrates. The energy difference (more precisely - enthalpy difference) was taken from surroundings (e.g. solvent), so the surroundings energy is lower. Thus, its temperature, which is a measure of average kinetic energy of molecules, is lower.

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By definition, an exothermic reaction gives off heat, and endothermic reaction (e.g. dissolution of ammonium nitrate in water, as in "cool-packs") absorbs heat from the environment in order to proceed.

Your assumption is based on the idea that heat leaves the reactants in order to cool them, which is not the case in an endothermic reaction, but is the case with a heat pump. These are different mechanisms; in the endothermic reaction, heat energy is used to force the reaction to proceed, e.g. a change of state from solid to liquid.

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You are absolutely right. And also wrong. You are right, when something becomes cooler, heat is released from it. Your mistake is you are looking for your answer from the wrong perspective, I believe.

When you are dealing with calorimetry and heat of reaction, it can seem contradictory at first. You learn that an exothermic reaction is one in which heat leaves, and an endothermic reaction is one in which heat enters the system. Then you go to determine if a process is endo or exo, and it seems backwards. When measuring, say a chemical reaction in water, you are determining whether the change was exothermic or endothermic. It can seem like the answer is exothermic when heat leaves the water. And it is- for the water. You want to determine the change for the chemical reaction. So, if water temperature decreases, then heat has left the water. For the water, this would be exothermic. But for the reaction, heat has entered that system. So the reaction is then, of course, endothermic.

Remember, you are determining heat of the system. Not heat of surroundings. System = chemicals Surroundings = Water

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What is your "system" in this case?

The reaction that you are performing--your reactants and your products--constitute your system. Since the reaction is endothermic, heat has to enter your system for the reaction to proceed.

But where does this heat come from? It is obviously drawn from the surroundings, and thus you measure a temperature that is now lower than what you started with.

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In an endothermic change, temperature is absorbed from surrounding molecules to continue reacting. If these molecules are losing heat, that means their temperature will drop, resulting in a temperature decrease.

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  • $\begingroup$ If you place a thermometer into a beaker where an exothermic reaction is reacting ( say, sodium hydroxide and water), the temperature will rise. That means that there is heat being released from the reaction? Isn't the inside of the beaker the location of the reactants themselves and shouldn't that temperature be lower because those molecules are releasing heat? $\endgroup$ – suse Oct 23 '18 at 3:47

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